Ralph wants to show 13/20

Vaimiti walks at a constant pace of 1.1 m/s and takes 25 minutes to get to school.

Temka [501]

Solution:

Note that:

Speed = Distance/Time
Vaimiti speed = 1.1 m/s
Jabril speed = 1.3 m/s

Converting the time (minutes to seconds) for Vaimiti to reach school:

Vaimiti's time to reach school: 25 minutes = 25 x 60 seconds
=> Vaimiti's time to reach school: 1500 seconds

Converting the time (minutes to seconds) for Jabril to reach school:

Jabril's time to reach school: 30 minutes = 30 x 60 seconds
=> Jabril's time to reach school: 1800 seconds

Finding the distance of Vaimiti:

Speed = Distance/Time

Important: The distance will be in meters since the speed units is meters/seconds.

=> 1.1 meters/second = Distance/1500
=> 1.1 x 1500 = Distance
=> 1650 meters = Distance (In meters)

Finding the distance of Jabril:

Speed = Distance/Time

Important: The distance will be in meters since the speed units is meters/seconds.

1.3 meters/second = Distance (In meters)/1800 seconds
=> 1.3 x 1800 = Distance (In meters)
=> 2340 meters = Distance (In meters)

This can lead to two possible solutions:

Possible solution #1:

Finding the difference between the two distances:

2340 meters - 1650 meters = Difference (In meters)
=> 690 meters

Possible solution #2:

The difference between the distances they walked is that Jabril walked faster than Vaimiti, but Vaimiti reached school earlier than Jabril because the walking distance for Vaimiti is less than the walking distance for Jabril.

Hoped this helped!

6 0

3 years ago

Read 2 more answers

2= 7/9 x 6 + b solve for b

netineya [11]

Answer:

8/3

Step-by-step explanation:

7/9 x 6

14/3

14/3 - 2

8/3

5 0

3 years ago

Find the lowest common denominator of and . A. (p + 2) B. (p + 5)(p + 3) C. (p + 5)(p + 2)(p + 3) D. (p + 2)2

laiz [17]

The lowest (or least) common denominator also written as LCD is the smallest of all the possible common denominators, where the denominator is the bottom number in a fraction.
We should find the lowest common denominator of (p+3)/(p^2+7p+10) and (p+5)/(p^2+5p+6).
p^2+7p+10 can be written as a product: (p+5)(p+2)
p^2+5p+6 can be written as a product: (p+3)(p+2)
So, we should find the LCD for (p+5)(p+2) and (p+3)(p+2). The smallest possible number that can be divided with both of them is:(p + 5)(p + 2)(p + 3)
Solution C.

6 0

3 years ago

2 parallelograms. The pre-image is smaller than the image.

statuscvo [17]

Answer: The figure changed size.

Step-by-step explanation: Dilations change the sizes of shapes. They can either make them bigger or make them smaller.

4 0

2 years ago

Consider the sequence: 4, 7, 10, ... , ... , .... What is the nth term of this sequence?

Tju [1.3M]

4+3=7+3=10+3=13. Steps.........

Ninth term in the sequence is 30